Question: Let $h(x)=x^4-2x^2$. What is the absolute maximum value of $h$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $-1$ (Choice B) B $0$ (Choice C) C $1$ (Choice D) D $h$ has no maximum value
Solution: Let's first find the relative extremum points of $h$, and then consider them along with the function's end behavior in both directions. We start with finding the critical points of $h$. The derivative of $h$ is $h'(x)=4x(x-1)(x+1)$. $h'(x)=0$ for $x=-1,0,1$. $h'$ is defined for all real numbers. Therefore, our critical points are $x=-1$, $x=0$, and $x=1$. Our critical points divide the function's domain (which is all real numbers) into four intervals: $\llap{-}2$ $\llap{-}1$ $0$ $1$ $2$ $x< \llap{-}1$ $\llap{-}1<x<0$ $0<x<1$ $x>1$ Let's evaluate $h'$ at each interval to see if it's positive or negative on that interval. Interval $x$ -value $h'(x)$ Verdict $x<-1$ $x=-2$ $h'\left(-2\right)=-24<0$ $h$ is decreasing $\searrow$ $-1<x<0$ $x=-\dfrac12$ $h'\left(-\dfrac{1}{2}\right)=\dfrac{3}{2}>0$ $h$ is increasing $\nearrow$ $0<x<1$ $x=\dfrac12$ $h'\left(\dfrac12\right)=-\dfrac32<0$ $h$ is decreasing $\searrow$ $x>1$ $x=2$ $h'\left(4\right)=24>0$ $h$ is increasing $\nearrow$ Now let's look at all the critical points: $x$ $h(x)$ Before After Verdict $-1$ $-1$ $\searrow$ $\nearrow$ Minimum $0$ $0$ $\nearrow$ $\searrow$ Maximum $1$ $-1$ $\searrow$ $\nearrow$ Minimum Let's imagine ourselves walking on the graph of $h$, starting all the way to the left (from $-\infty$ ) and going all the way to the right (until $+\infty$ ). According to the table, we will start by going down until $(-1,-1)$, then go up until $(0,0)$, then down again until $(1,-1)$, and then forever go up. This means that $\lim_{x\to-\infty}h(x)=\lim_{x\to +\infty}h(x)=+\infty$, which means $h$ has no maximum value. In conclusion, $h$ has no absolute maximum value.